Problem: Graph this system of equations and solve. $4x+4y = -8$ $8x+2y = 8$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Solution: Convert the first equation, $4x+4y = -8$ , to slope-intercept form. $y = - x - 2$ The y-intercept for the first equation is $-2$ , so the first line must pass through the point $(0, -2)$ The slope for the first equation is $-1$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move down (because it's negative) You must also move $1$ position to the right. $1$ position to the right. $1$ position down from $(0, -2)$ is $(1, -3)$ Graph the blue line so it passes through $(0, -2)$ and $(1, -3)$ Convert the second equation, $8x+2y = 8$ , to slope-intercept form. $y = -4 x + 4$ The y-intercept for the second equation is $4$ , so the second line must pass through the point $(0, 4)$ The slope for the second equation is $-4$ . Remember that the slope tells you rise over run. So in this case for every $4$ positions you move down (because it's negative) $1$ position to the right. $4$ positions down from $(0, 4)$ is $(1, 0)$ Graph the green line so it passes through $(0, 4)$ and $(1, 0)$ The solution is the point where the two lines intersect. The lines intersect at $(2, -4)$.